/*
Delete Node in a BST
====================

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.

Basically, the deletion can be divided into two stages:

Search for a node to remove.
If the node is found, delete the node.
Follow up: Can you solve it with time complexity O(height of tree)?

Example 1:
Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: Given key to delete is 3. So we find the node with value 3 and delete it.
One valid answer is [5,4,6,2,null,null,7], shown in the above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.

Example 2:
Input: root = [5,3,6,2,4,null,7], key = 0
Output: [5,3,6,2,4,null,7]
Explanation: The tree does not contain a node with value = 0.

Example 3:
Input: root = [], key = 0
Output: []

Constraints:
The number of nodes in the tree is in the range [0, 104].
-105 <= Node.val <= 105
Each node has a unique value.
root is a valid binary search tree.
-105 <= key <= 105
*/

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

class Solution
{
public:
  TreeNode *deleteNode(TreeNode *root, int key)
  {
    if (!root)
      return root;
    else if (root->val > key)
      root->left = deleteNode(root->left, key);
    else if (root->val < key)
      root->right = deleteNode(root->right, key);
    else
    {
      if (!root->left)
        return root->right;
      if (!root->right)
        return root->left;

      auto curr = root->right;
      while (curr->left)
        curr = curr->left;

      root->val = curr->val;
      root->right = deleteNode(root->right, root->val);
    }

    return root;
  }
};